Chapter 4, Problem 4.4P

To determine

The shredder energy requirement for shredding.

Answer to Problem 4.4P

The shredder energy requirement for shredding is $7.61\text{kWh}/\text{ton}$ .

Explanation of Solution

Given:

The fraction of feed retained on sieve is shown in the table below:

Sieve Size $\left(\text{inches}\right)$	Fraction of Feed Retained on the Sieve
$10$	$0.05$
$8$	$0.20$
$4$	$0.32$
$2$	$0.18$
$1$	$0.20$

Assume, from the figure that the axis is a logarithmic scale then for $Y$ at $0.62$ and $0.1$ , the size of the product is $0.8\text{in,}$ and the size of the feed is $4\text{in}$ .

Calculation:

Write the equation to calculate the shredder energy requirement for shredding.

$E=\frac{10E_i}{{\left[x_0{\left(1.61\right)}^{\frac{1}{n}}\right]}^{\frac{1}{2}}}-\frac{10E_i}{L_f{}^{\frac{1}{2}}}$ ...... (I)

Here, the shredder energy is $E$ , the work index is $E_i$ , the constant is $n$ , the characteristic size is $x_0,$ and the length of the feed is $L_f$ .

Write the equation to calculate the slope $n_p$ .

$n_p=\frac{Y_1-Y_2}{x_{0p}-x_p}$ ...... (II)

Substitute $0.62$ for $Y_1$ , $0.1$ for $Y_2$ , $0.8$ for $x_{0p},$ and $0.08$ for $x_p$ in Equation (II).

$\begin{array}{c}{n}_p=\frac{0.62-0.1}{0.8-0.08}\\ =\frac{0.52}{0.72}\\ =0.72\end{array}$

Write the equation to calculate the slope $n_s$ .

$n_s=\frac{Y_1-Y_2}{x_{0f}-x_f}$ ...... (III)

Substitute $0.62$ for $Y_1$ , $0.1$ for $Y_2$ , $4$ for $x_{0f},$ and $2$ for $x_f$ in Equation (III).

$\begin{array}{c}{n}_s=\frac{0.62-0.1}{4-2}\\ =\frac{0.52}{2}\\ =0.26\end{array}$

The slope $n_p$ is greater.

Calculate the shredder energy requirement for shredding.

Substitute $0.72$ for $n$ , $4\text{in}$ for $L_f$ , $0.8\text{in}$ for $x_0,$ and $400\text{kWh}/\text{ton}$ in the Equation (I).

$\begin{array}{c}E=\frac{10\left(400\text{kWh}/\text{ton}\right)}{{\left[0.8\text{in}\times \left(\frac{25400\text{μm}}{1\text{in}}\right){\left(1.61\right)}^{\frac{1}{0.72}}\right]}^{\frac{1}{2}}}-\frac{10\left(400\text{kWh}/\text{ton}\right)}{{\left[4\text{in}\times \left(\frac{25400\text{μm}}{1\text{in}}\right)\right]}^{\frac{1}{2}}}\\ =\frac{10\left(400\text{kWh}/\text{ton}\right)}{198.4}-\frac{10\left(400\text{kWh}/\text{ton}\right)}{318.74}\\ =20.16\text{kWh}/\text{ton}-12.55\text{kWh}/\text{ton}\\ =7.61\text{kWh}/\text{ton}\end{array}$

Conclusion:

Thus, the shredder energy requirement for shredding is $7.61\text{kWh}/\text{ton}$ .